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The graph of f is shown. Evaluate each integral by interpreting it in terms of areas.

The main objective of this question is to find the area under the curve by evaluating the given integral.

This question uses the concept of Integral. Integrals can be used to find the area of the given expression under the curve by evaluating it.

Expert Answer

We have to find the area by evaluating the integral. We are given with:

\[ \int_{0}^{2}  f(x) \,dx \]

We first divided the area into two parts. In the first part, we have to find the area of the triangle which is:

\[= \space \frac{1}{2}Base . Height  \]

By putting values in the above equation, we get:

\[= \space \frac{1}{2} 2 . 2 \]

\[= \space \frac{1}{2} 4 \]

Dividing  $ 4 $ by $ 2 $ results in:

\[= \space 2 \]

So, the area of a triangle is $ 2 $.

Now we have to calculate the area of the square which is:

\[ \int_{0}^{2}  f(x) \,dx \]

\[=\space 2 \space + \space 2 \]

\[= \space 4]

So the area of the square is $ 4 $ units.

Numerical Results

The area of the given integral under the curve is $ 2 $ and $ 4 $ units.

Example

Find the area of the given integral in the graph.

  1. \[ \int_{0}^{20}  f(x) \,dx \]
  2. \[ \int_{0}^{50}  f(x) \,dx \]
  3. \[ \int_{50}^{70}  f(x) \,dx \]

We have to find the area of the given integrals by evaluating them.

First, we will find the area for the limit 0 to 20. Area is :

\[10 \space \times \space 20 \space + \space  \frac{1}{2} \times 20 \times 20 \]

\[200 \space + \space  \frac{1}{2} \times 20 \times 20 \]

\[200 \space + \space   10 \times 20 \]

\[200 \space + \space   200 \]

\[400  units\]

Now, we have find the area for the limit $ 0 $ to  $ 50 $. Area is :

\[10 \space \times \space 30 \space + \space  \frac{1}{2} \times 30 \times 20 \]

\[300 \space + \space  \frac{1}{2} \times 30 \times 20 \]

\[300 \space + \space   30 \times 10 \]

\[300 \space + \space   300 \]

\[600  units\]

Now for the limit of $ 50 $ to $ 70 $ , the area is:

\[=\space \frac{1}{2} (-30) (20) \]

\[= – 300 \]

Now for the limit of $ 0 $ to $ 90 $, the area is:

\[= \space 400 \space + \space 600 \space – \space 300 \space – \space 500 \]

\[= \space 200 units \]

The area for the given integrals is $ 400 $, $ 1000$ , $ 300 $ , and $ 200 $ units.

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